To better understand certain problems involving
aircraft and propulsion it is necessary to use some
mathematical ideas from trigonometry
the study of triangles. Let us begin with some
definitions and terminology which we will use on this
slide. We start with a right triangle. A right
triangle is a three sided figure with one angle equal
to 90 degrees. A 90 degree angle is called a right
angle and that is where the right triangle gets
its name. We define the side of the triangle opposite
from the right angle to be the hypotenuse, h.
It is the longest side of the three sides of the right
triangle. The word "hypotenuse" comes
from two Greek words meaning "to stretch",
since this is the longest side. We are going to
label the other two sides a and b. The Pythagorean
Theorem is a statement relating the lengths of the
sides of any right triangle.

The theorem states that:

For any right triangle, the square of the
hypotenuse
is equal to the sum of the squares of the other two
sides.

Mathematically, this is written:

h^2 = a^2 + b^2

The theorem has been known in many cultures, by
many names, for many years. Pythagoras, for whom the
theorem is named, lived in ancient Greece, 2500 years
ago. It is believed that he learned the theorem during
his studies in Egypt. The Egyptians probably knew of
the relationship for a thousand years before
Pythagoras. The Egyptians knew of this relationship
for a triangle with sides in the ratio of "3 - 4
- 5".

5^2 = 3^2 + 4^2

25 = 9 + 16

Pythagoras generalized the result to any right
triangle. There are many different algebraic and
geometric proofs of the theorem. Most of these
begin with a construction of squares on a sketch of a
basic right triangle. On the figure at the top of this
page, we show squares drawn on the three sides of the
triangle. A square is the special case of a rectangle
in which all the sides are equal in length. The areaA of a rectangle is the product of the sides.
So for a square with a side equal to a, the
area is given by:

A = a * a = a^2

So the Pythagorean theorem states the area h^2
of the square drawn on the hypotenuse is equal to the
area a^2 of the square drawn on side a
plus the area b^2 of the square drawn on side b.

Here's an interactive Java program that let's you
see that this area relationship is true:

We begin with a right triangle on which we have
constructed squares on the two sides, one red and one
blue. We are going to break up the pieces of these two
squares and move them into the grey square area on the
hypotenuse. We won't loose any material during the
operation. So if we can exactly fill up the square on
the hypotenuse, we have shown that the areas are
equal. You work through the construction by clicking
on the button labeled "Next". You can go
"Back" and repeat a section, or go all the
way back to the beginning by clicking on
"Reset".

What is it doing? The first step rotates the
triangle down onto the blue square. This cuts the blue
square into three pieces, two triangles and a red
rectangle. The two triangles are exactly the same size
as the original triangle. The "bottom" of
the original triangle exactly fits the vertical side
of the square, because the sides of a square are
equal. The red rectangle has its vertical sides equal
to the base of the original triangle, and its
horizontal sides equal to the difference between the
"bottom" side and the "vertical"
side of the original triangle. Using the terminology
from the figure at the top of this page, the
dimensions of the red rectangle are:

vertical length = b

horizontal length = b - a

The next step is to move the red rectangle over
adjacent to the red square. The rectangle sticks out
the top of the red square and the two triangles remain
in the blue square. The next step is to move one of
the blue triangles vertically into the hypotenuse
square. It fits exactly along the side of the
hypotenuse square because the sides of a square are
equal. The next step is to move the other blue
triangle into the hypotenuse square. (We are half way
there!) The next step is to slide the form of the
original triangle to the left into the red region. The
triangle cuts the red region into three pieces, two
triangles and a small yellow square. The original
triangle fits exactly into this region because of two
reasons; the vertical sides are identical, and the
horizontal side of the red region is equal to the
length of the red square plus the horizontal length of
the red rectangle which we moved. The horizontal
length of the red region is:

horizontal length = a + (b - a) = b

The horizontal length of the red region is exactly
the length of the horizontal side of the original
triangle. The yellow square has dimensions b - a
on each side. The next step is to move one of the red
triangles into the hypotenuse square. Again it's a
perfect fit. The next step is to move the final red
triangle into the hypotenuse square. Now if we look at
the grey square that remains in the hypotenuse square,
we see that its dimensions are b - a; the long
side of the triangle minus the short side. The final
step is to move the yellow square into this hole. It's
a perfect fit and we have used all the material from
the original red and blue squares.